Numpy Vector (N,1) dimension -> (N,) dimension conversion

Question

I have a question regarding the conversion between (N,) dimension arrays and (N,1) dimension arrays. For example, y is (2,) dimension.

A=np.array([[1,2],[3,4         


        

Answer 1

Slice along the dimension you want, as in the example below. To go in the reverse direction, you can use None as the slice for any dimension that should be treated as a singleton dimension, but which is needed to make shapes work.

In [786]: yy = np.asarray([[11],[7]])

In [787]: yy
Out[787]:
array([[11],
       [7]])

In [788]: yy.shape
Out[788]: (2, 1)

In [789]: yy[:,0]
Out[789]: array([11, 7])

In [790]: yy[:,0].shape
Out[790]: (2,)

In [791]: y1 = yy[:,0]

In [792]: y1.shape
Out[792]: (2,)

In [793]: y1[:,None]
Out[793]:
array([[11],
       [7]])

In [794]: y1[:,None].shape
Out[794]: (2, 1)

Alternatively, you can use reshape:

In [795]: yy.reshape((2,))
Out[795]: array([11,  7])

Answer 2

What about vice versa? Numpy Numpy Vector (N,) dimension conversion ->Vector (N,1) dimension dimension conversion

Answer 3

the opposite translation can be made by:

np.atleast_2d(y).T

Answer 4

reshape works for this

a  = np.arange(3)        # a.shape  = (3,)
b  = a.reshape((3,1))    # b.shape  = (3,1)
b2 = a.reshape((-1,1))   # b2.shape = (3,1)
c  = b.reshape((3,))     # c.shape  = (3,)
c2 = b.reshape((-1,))    # c2.shape = (3,)

note also that reshape doesn't copy the data unless it needs to for the new shape (which it doesn't need to do here):

a.__array_interface__['data']   # (22356720, False)
b.__array_interface__['data']   # (22356720, False)
c.__array_interface__['data']   # (22356720, False)

Answer 5

Use numpy.squeeze:

>>> x = np.array([[[0], [1], [2]]])
>>> x.shape
(1, 3, 1)
>>> np.squeeze(x).shape
(3,)
>>> np.squeeze(x, axis=(2,)).shape
(1, 3)